Metric completion of universal covering of punctured plane It is known that the universal covering of the punctured plane $\mathbb C\setminus\{0\}$ is $\exp:\mathbb C\to\mathbb C\setminus\{0\}$. In real coordinates,
$f=\exp:\tilde M=\mathbb R^2\to M=\mathbb R^2\setminus\{(0,0)\}$ is given by 
$f(x,y)=(e^x\cos y,e^x\sin y)$. The Euclidean metric on $M$ lifts to a flat metric on $\tilde M$ which is not complete (for otherwise $M$ would be complete). 
We can also identify $\tilde M$ with $\mathbb R^+\times\mathbb R$ via the map
$(x,y)\mapsto(e^x,y)$, and then the lifted metric is $dx^2+x^2\,dy^2$. My question is

What is the metric completion $\hat M$ of the universal covering of the punctured plane?

It seems there are Cauchy sequences $\{(\frac 1n,y_0)\}_n$ in $\tilde M$ which define the same point of $\hat M$ for all $y_0\in\mathbb R$. 
 A: Let $g = dx^2 + x^2dy^2$. First observe that  for $(x,y)$,$(a,b) \in (\mathbb R^+ \times \mathbb R,g)$ for all $n \geq 1:$
 $$d((x,y),(a,b)) \leq d((1/n,y),(x,y)) + d((1/n,y),(1/n,b)) + d((1/n,b),(a,b))$$
$$\leq x + 1/n|y - b| + a.$$
It follows that 
$$d((x,y),(a,b)) \leq x + a.$$
Now let $p_n = (x_n,y_n)$ be a cauchy sequence in $\mathbb R^+ \times \mathbb R$ with respect to $g$. Consider the local isometry $\pi : \mathbb R^+ \times \mathbb R \to \mathbb R^2 \setminus \{0\}$, where the second space is equipped with the flat standard metric. It is easy to see that $\pi(p_n)$ is a cauchy sequence as well and $p_n$ converges if and only if $\pi(p_n)$ does. Thus if $p_n$ does not converge with respect to $g$ it follows that $||\pi(p_n)||$ is bounded. On the other hand $\pi(p_n)$ is a divergent cauchy sequence, so it follows that $\pi(p_n)$, considered as a sequence in $\mathbb R^2$, converges to $0$. Equivalently $x_n \to 0$ for $n \to \infty$. We have shown
A cauchy sequence $p_n = (x_n,y_n)$ in $(\mathbb R^+ \times \mathbb R,g)$ does not converge if and only if $x_n \to 0$.
Now let $q_n = (a_n,b_n)$ be another cauchy sequence that does not converge in ($\mathbb R^+ \times \mathbb R,g)$. So $a_n \to 0$. It follows from the above inequality that $d((x_n,y_n),(a_n,b_n)) \leq |x_n - a_n| \to 0$ for $n \to \infty$.
From the construction of the completion of $(\mathbb R^+ \times \mathbb R,d)$ it follows, that it is isometric to $(\mathbb R^+ \times \mathbb R \cup \{pt\},\hat d)$, where $\{pt\}$ is a point disjoint from $\mathbb R^+ \times \mathbb R$ and the metric is given by $\hat d((x,y),(a,b)) = d((x,y),(a,b))$, $\hat d((x,y),pt) = x$ and $\hat d(pt,pt) = 0$. Intuitively $\hat d$ is constructed by collapsing the $y$ axis to a point, which corresponds to all nonconvergent cauchy sequences.
